If the line $\dfrac{1-x}{-3}=y=\dfrac{z+2}{2}$ is perpendicular to the line $\dfrac{3 x-1}{2 b}=3-y=\dfrac{z-1}{a}$, then find the value of $3 a+3 b$

The first line is given by $\dfrac{x-1}{3} = \dfrac{y}{1} = \dfrac{z+2}{2}$. The direction vector of this line is $\vec{v_1} = 3\hat{i} + 1\hat{j} + 2\hat{k}$.

The second line is given by $\dfrac{3x-1}{2b} = \dfrac{y-3}{-1} = \dfrac{z-1}{a}$. Rewriting this in standard form $\dfrac{x - 1/3}{2b/3} = \dfrac{y-3}{-1} = \dfrac{z-1}{a}$. The direction vector of this line is $\vec{v_2} = \dfrac{2b}{3}\hat{i} - 1\hat{j} + a\hat{k}$.

Since the lines are perpendicular, the dot product of their direction vectors must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.

$(3)\left(\dfrac{2b}{3}\right) + (1)(-1) + (2)(a) = 0$.

$2b - 1 + 2a = 0$.

$2a + 2b = 1$.

We need to find the value of $3a + 3b$.

$3(a + b) = 3 \times \dfrac{1}{2} = \dfrac{3}{2}$.

Answer: $\dfrac{3}{2}$